Can you prove this...?

Question

n > 6 and an integer

Prove that if n + 1 and n - 1 are both prime, n^2(n^2+16) is divisible by 720.

Is the converse true, i.e. if n^2(n^2+16) is divisible by 720, are both n + 1 and n - 1 prime?

As babeladitya remarks, this is indeed a question from an old maths competition that I've been trying to solve.

Answer 1

Suppoe that n+1 and n-1 are prime, and n>6. Then n+1 ≠ 0 mod 5 and n-1 ≠ 0 mod 5 (Note: the ≠ sign should be a "not congruent to" sign, but the proper symbol won't show up here). Therefore, n≠1 mod 5 and n≠4 mod 5. Therefore, n≡0, 2, or 3 mod 5. If n≡0 mod 5 then n²(n²+16)≡0 mod 5. If n≡2 mod 5 then n²(n²+16)≡4(4+1)≡0 mod 5. Finally, if n≡3 mod 5, n²(n²+16)≡4(4+1)≡0 mod 5. Therefore, in any case, n²(n²+16)≡0 mod 5 and n²(n²+16) is divisible by 5.

Now note that n is divisible by both 2 and 3. (proof: n+1 and n-1 are both prime, and therefore odd, and therefore n is even. Similarly, if n≡1 mod 3 then n-1 is divisible by three and not prime, and if n≡2 mod 3 then n+1 is not prime, therefore n≡0 mod 3). Since n is divisible by 3, it follows that n² is divisible by 9, and n²(n²+16), being a multiple of n², is also divisible by 9.

Now note that n²(n²+16) = n⁴+16n². Now, since n is divisible by 2, it follows that n⁴ is divisible by 16, and 16n² is clearly also divisble by 16, thus the sum must be divisible by 16.

Having established that n²(n²+16) is divisible by 5, 9 and 16, it follows that it must be divisible by the least common multiple of 5, 9, and 16, which is 720. Q.E.D.

Converse: this is false. Let n=48. Then n²(n²+16) = 5,345,280, which is 7424*720, however n+1=49=7*7, which is not prime.

Answer 2

n must be even, or else n+1 and n-1 are even and larger than 2 so that they are not prime. Note that 720=6*5*4*3*2=16*9*5. So we have to show that the number is divisible by 16, 9, and 5.

Write n as 2k for some integer k. n^2(n^2+16) = (2k)^2((2k)^2+16) = 16k^2(k^2+4). So the number is divisible by 16.

Since n+1 and n-1 are prime and n>6, the last digit of n cannot be 4 or 6. Otherwise, the last digit of n+1 or n-1 would be 5 but the number would not be equal to 5. But this means that n+1 or n-1 is not prime, since numbers ending in 5 are divisible by 5. If the last digit of n is 0, then n is divisible by 5. So n^2(n^2+16) is also divisible by 5. If the last digit of n is 2, then the last digit of n^2 is 4 and the last digit of n^2+16 is 0. Again, this means n^2(n^2+16) is divisible by 5. If the last digit of n is 8, then the last digit of n^2 is 4 and the last digit of n^2+16 is 0. Again, this means n^2(n^2+16) is divisible by 5. These are all cases of last digit, so we conclude that n^2(n^2+16) is divisible by 5.

If n is divisible by 3, then n^2 is divisible by 9. So n^2(n^2+16) is divisible by 9. If n has remainder 1 when divided by 3, then n-1 is divisible by 3 and not equal to 3, so it is not prime. (Recall that n>6.) If n has remainder 2 when divided by 3, then n+1 is divisible by 3 and not equal to 3, so it is not prime. (Again, n>6.) Hence, n is divisible by 3 so that n^2(n^2+16) is divisible by 9.

This proves that n^2(n^2+16) is divisible by 720.

The converse fails, however. Take n=48. 48^2(48^2+16) = 29*5*(3^2)*(2^12)= 720*7424, but n+1 = 49 = 7*7 is not prime.

Answer 3

The converse is false. The sequence 5*3*2^(2k+1) for nonnegative k's gives numbers (call them n) such that n^2 (hence n^2(n^2+16) too) is divisible by 5*3^2*2^4=6!=720, and the claim would imply n-1 and n+1 are both primes, so we could construct an infinite sequence of primes (even, pairs of twin primes!), which is notoriously impossible. If you want an explicit counterexample, just take k=1, i.e. n=5*3*8=120, but n+1=121=11^2.

Answer 4

For the first part, try letting n = 2m since n must be even.

The converse is not true. Let n = 720, then
n^2 (n^2 +16) is divisible by 720, but
720+1 = 721 is not prime (721 = 7 times103)

Answer 5

i did this question for a contest. i remember a few years ago.
british mathematical olympiad, was it? that competition is not easy, but it sure is fun!


just check out the b.m.o. website ... i cannot remember the answer off the top of my head ... and do not really have time to work it all out.

however, i can assure you that you will enjoy doing the problem. so if u're any good at math, give it a go.


here's another problem for any geniuses out there:
find the smallest arithmetic sequence of 7 positive primes.
(for example, an arithmetic sequence of 3 positive primes would be 7, 13, 19)

I'll give you people a quick hint of where to start. i have proven that the difference between any two successive primes has to be a multiple of 30 (i.e. 7, 37, 67 ... or 7, 67, 127 ... or even maybe 7, 127, 247 .... . enjoy!

Answer 6

I cant. Havent taken number theory yet.

There is a conjecture about paired primes. Although its a weak support, if such a simple test for paired primes were possible, the conjecture would be provable. I think the converse is untrue.

I was able to test the first statement, and its accurate up to 19542. I dont know if the mod test starts doing roundoff then or not.

Answer 7

dependant on the value of n there is one instance where both the equation and its converse are both true.

Answer 8

AS Farmer Jones said to his hired hand ... there is a lot of bullocks in the field .....

Answer 9

maybe later

Answer 10

dont be stupid its the other way round, god dont you know anything